cee 271: applied mechanics ii, dynamics lecture 2: ch.12, sec.3h 5 · 2013. 9. 11. · 12.3...

12.3 Rectilinear Motion (cont’d)

CEE 271: Applied Mechanics II, Dynamics– Lecture 2: Ch.12, Sec.3h–5 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

Date: __________________

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RECTILINEAR KINEMATICS: ERRATIC MOTION(Section 12.3)

Today’s objectives: Studentswill be able to

1 Determine position,velocity, and acceleration ofa particle using graphs.

In-class activities:• Reading Quiz• Applications• s− t, v − t, a− t, v − s,

and a− s diagrams• Concept Quiz• Group Problem Solving• Attention Quiz

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EXAMPLE

• Given: The s− t graph for a sports car moving along astraight road.

• Find: The v − t graph and a− t graph over the time intervalshown.

• What is your plan of attack for the problem?3 / 30


EXAMPLE (continued)

• Solution: The v − t graph can be constructed by finding theslope of the s− t graph at key points. What are those?

• When 0 ≤ t ≤ 5 s ;

v0−5 =ds

dt=d(3t2)

dt= 6tm/s

• When 5 ≤ t ≤ 10 s;

v5−10 =ds

dt=d(30t− 75)

dt= 30 m/s

v(m/s)

t(s)

30

5 10

v-t graph

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EXAMPLE (continued)

• Similarly, the a− t graph can be constructed by finding theslope at various points along the v − t graph.

• When 0 ≤ t ≤ 5 s;

a0−5 =dv

dt=d(6t)

dt= 6 m/s2

• When 5 ≤ t ≤ 10 s;

a5−10 =dv

dt=d(30)

dt= 0 m/s2

a(m/s2)

t(s)

6

5 10

a-t graph

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CONCEPT QUIZ

1 If a particle starts from rest andaccelerates according to the graphshown, the particle’s velocity att = 20 s is(A) 200 m/s(B) 100 m/s(C) 0(D) 20 m/s

ANS: (B)2 The particle in Problem 1 stops

moving at t =

(A) 10 s(B) 20 s(C) 30 s(D) 40 s

ANS: (D)

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GROUP PROBLEM SOLVING

• Given: The v − t graph shown.• Find: The a− t graph, average speed, and distance

traveled for the 0 - 90 s interval.• Plan: Find slopes of the v − t curve and draw the a− t

graph. Find the area under the curve. It is the distancetraveled. Finally, calculate average speed (using basicdefinitions!).

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GROUP PROBLEM SOLVING (continued)

-0.5

1

a(m/s²)

30 90 t(s)

a-t graph

• Solution:• Find: The a− t graph:• For 0 ≤ t ≤ 30,

a =dv

dt= 1.0 m/s2

• For 30 ≤ t ≤ 90,

a =dv

dt= −0.5 m/s2

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• Now find the distance traveled:

∆s0−30 =

∫vdt =

[12 t

2]300

= 450 m (1)

∆s30−90 =

∫vdt

=[12(−0.5)t2 + 45t

]9030

= 900 m (2)s0−90 = 450 + 900 = 1350 m (3)

vavg(0−90) = total distance / time = 1350/90 = 15 m/s (4)

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ATTENTION QUIZ 1

t

v

6 s

75

1. If a car has the velocity curve shown, determine the time tnecessary for the car to travel 100 meters.(A) 8 s(B) 4 s(C) 10 s(D) 6 s

ANS: (B)

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ATTENTION QUIZ 2

t

v

2. Select the correct a− t graph for the velocity curve shown.

A) B)

C) D)

a

t

a

t

a

t

a

t

ANS: (D)11 / 30


CURVILINEAR MOTION: GENERAL &RECTANGULAR COMPONENTS

Today’s objectives: Studentswill be able to

1 Describe the motion of aparticle traveling along acurved path.

2 Relate kinematic quantitiesin terms of the rectangularcomponents of the vectors.

In-class activities:• Reading Quiz• Applications• General Curvilinear Motion• Rectangular Components

of Kinematic Vectors• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 In curvilinear motion, the direction of the instantaneousvelocity is always(a) tangent to the hodograph.(b) perpendicular to the hodograph.(c) tangent to the path.(d) perpendicular to the path.

ANS: (C)

2 In curvilinear motion, the direction of the instantaneousacceleration is always(a) tangent to the hodograph.(b) perpendicular to the hodograph.(c) tangent to the path.(d) perpendicular to the path.

ANS: (A)

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APPLICATIONS

• The path of motion of aplane can be tracked withradar and its x, y, and zcoordinates (relative to apoint on earth) recorded asa function of time.

• How can we determine thevelocity or acceleration ofthe plane at any instant?

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APPLICATIONS(continued)

• A roller coaster car travels down a fixed, helical path at aconstant speed.

• How can we determine its position or acceleration at anyinstant?

• If you are designing the track, why is it important to be ableto predict the acceleration of the car?

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GENERAL CURVILINEAR MOTION (Section 12.4)

• A particle moving along a curved path undergoescurvilinear motion. Since the motion is oftenthree-dimensional, vectors are used to describe themotion. A particle moves along a curve defined by thepath function, s.

• The position of the particle at any instant is designated bythe vector r = r(t). Both the magnitude and direction of rmay vary with time.

• If the particle moves a distance ∆s along the curve duringtime interval ∆t, the displacement is determined by vectorsubtraction: ∆r = r′ − r 16 / 30


VELOCITY

• Velocity represents the rate ofchange in the position of a particle.

• The average velocity of the particleduring the time increment ∆t isvavg = ∆r/∆t.

• The instantaneous velocity is thetime-derivative of position v = dr/dt.

• The velocity vector, v, is always tangent to the path ofmotion.

• The magnitude of v is called the speed. Since the arclength ∆s approaches the magnitude of ∆r as t→ 0, thespeed can be obtained by differentiating the path function(v = ds/dt). Note that this is not a vector!

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ACCELERATION

• Acceleration represents the rate of changein the velocity of a particle.

• If a particle’s velocity changes from v to v′

over a time increment ∆t, the averageacceleration during that increment is :aavg = ∆v/∆t = (v′ − v)/∆t

• The instantaneous acceleration is thetime-derivative of velocity :a = dv/dt = d2r/dt2

• A plot of the locus of points defined by thearrowhead of the velocity vector is called ahodograph. The acceleration vector istangent to the hodograph, but not, ingeneral, tangent to the path function.

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CURVILINEAR MOTION: RECTANGULARCOMPONENTS (Section 12.5)

• It is often convenient to describethe motion of a particle in termsof its x, y, z or rectangularcomponents, relative to a fixedframe of reference.

• The position of the particle canbe defined at any instant by theposition vector: r = xi + yj + zk.

• The x, y, z components may all be functions of time, i.e.,x = x(t), y = y(t), and z = z(t) .

• The magnitude of the position vector is: r = (x2 + y2 + z2)0.5

• The direction of r is defined by the unit vector: ur = r/r

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RECTANGULAR COMPONENTS: VELOCITY

• The velocity vector is the time derivative of the positionvector:

v =dr

dt=d(xi)

dt+d(yj)

dt+d(zk)

dt

• Since the unit vectors i, j, k are constant in magnitude anddirection, this equation reduces to

v = vxi + vyj + vzk

where

vx = x =dx

dtvy = y =

dy

dtvz = z =

dz

dt

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RECTANGULAR COMPONENTS: VELOCITY

• The magnitude of the velocity vector is

v =[(vx)2 + (vy)2 + (vz)

2]0.5

• The direction of v is tangent to the path of motion.

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RECTANGULAR COMPONENTS: ACCELERATION

• The acceleration vector is the time derivative of the velocityvector (second derivative of the position vector):

a =dv

dt=d2r

dt2= axi + ayj + azk

where ax = vx = x = dvx/dt, ay = vy = y = dvy/dt, andaz = vz = z = dvz/dt.

• The magnitude of the acceleration vector isa = [(ax)2 + (ay)2 + (az)

2]0.5 or a =√a · a.

• The direction of a is usually not tangent to the path of theparticle.

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EXAMPLE

• Given: The motion of two particles (A and B) is describedby the position vectors

rA = xAi + yAj = [3ti + 9t(2− t)j] m (5)rB = xBi + yBj = [3(t2 − 2t+ 2)i + 3(t− 2)j] m (6)

• Find: The point at which the particles collide and theirspeeds just before the collision.

• Plan:1 The particles will collide when their position vectors are

equal, or rA = rB .2 Their speeds can be determined by differentiating the

position vectors.

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EXAMPLE (continued)

Solution:1 The point of collision requires that rA = rB, so xA = xB

and yA = yB .• Set the x-components equal: xA = xB , 3t = 3(t2 − 2t+ 2)

• Simplifying: t2 − 3t+ 2 = 0• Solving: t = 3± [32 − 4(1)(2)]0.5/2(1) so that t = 2 or 1 s.

• Set the y-components equal: yA = yB , 9t(2− t) = 3(t− 2)

• Simplifying: 3t2 − 5t− 2 = 0• Solving: t = 5± [52 − 4(3)(−2)]0.5/2(3)

so that t = 2 or −1/3 s.

• So, the particles collide when t = 2 s (only common time).Substituting this value into rA or rB yields xA = xB = 6 mand yA = yB = 0.

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EXAMPLE (continued)From

rA = xAi + yAj = [3ti + 9t(2− t)j] mrB = xBi + yBj = [3(t2 − 2t+ 2)i + 3(t− 2)j] m

1 Differentiate rA and rB to get the velocity vectors.

vA =drAdt

= xAi + yAj = [3i + (18− 18t)j] m/s (7)

vB =drBdt

= xBi + yBj = [(6t− 6)i + 3j] m/s (8)

• At t = 2 s: vA = [3i− 18j] m/s and• At t = 2 s: vB = [6i + 3j] m/s

• Speed is the magnitude of the velocity vector.

|vA| = (32 + (−18)2)0.5 = 18.2 m/s (9)|vB | = (62 + 32)0.5 = 6.71 m/s (10)

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CHECK YOUR UNDERSTANDING QUIZ

1 If the position of a particle is defined byr = [(1.5t2 + 1)i + (4t− 1)j] (m), its speed at t = 1 s is(a) 2 m/s(b) 3 m/s(c) 5 m/s(d) 7 m/s

ANS: (c)

2 The path of a particle is defined by y = 0.5x2. If thecomponent of its velocity along the x-axis at x = 2 m isvx = 1 m/s, its velocity component along the y-axis at thisposition is(a) 0.25 m/s(b) 0.5 m/s(c) 1 m/s(d) 2 m/s

ANS: (d)

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GROUP PROBLEM SOLVING

• The velocity of the particle is

v(t) = [16t2i + 4t3j + (5t+ 2)k] m/s (11)= vxi + vyj + vzk (12)

and x = y = z = 0 at t = 0.• Find: The particle’s coordinate position and the magnitude

of its acceleration when t = 2 s.• Plan: Note that velocity vector is given as a function of

time.1 Determine the position and acceleration by integrating and

differentiating v, respectively, using the initial conditions.2 Determine the magnitude of the acceleration vector usingt = 2 s.

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Solution:1 x-components:

vx = x =dx

dt= (16t2) (13)∫ x

0dx = x =

∫ t

016t2dt =

[16

3t3]20

= 42.7 (14)

⇒ x = (16/3)t3 = 42.7 meter at t = 2 sAcceleration: ax = x = vx = d(16t2)

dt = 32t = 64 m/s2

2 y-components:Velocity known as : vy = y = dy

dt = 4t3 m/s

Position:∫ y0 dy =

∫ t0 4t3dt⇒ y =

[t4]20

= 16 meter at t = 2s

Acceleration: ay = y = vy = d(4t3)dt =

(12t2

)t=2

= 48m/s2

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Solution:3. z-components:

Velocity known as : vz = z = dzdt = (5t+ 2)m/s

Position:∫ z0 dz =

∫ t0 (5t+ 2) dt⇒ z =

[52 t

2 + 2t]20

= 14 m att = 2 sAcceleration: az = z = vz = d(5t+2)

dt = 5 m/s2

4. The position vector and magnitude of the accelerationvector are written using the component information foundabove.Position vector: r = [42.7i + 16j + 14k] m.Acceleration vector: a = [64i + 48j + 5k] m/s2

Magnitude: a = (642 + 482 + 52)0.5 = 80.2 m/s2

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ATTENTION QUIZ

1. If a particle has moved from A to Balong the circular path in 4s, what is theaverage velocity of the particle?(a) 2.5i m/s(b) 2.5i + 1.25j m/s(c) 1.25πi m/s(d) 1.25πj m/s

A B

R=5m x

y

ANS: (a)

2. The position of a particle is given as r = (4t2i− 2xj) m.Determine the particle’s acceleration.(a) (4i + 8j) m/s2

(b) (8i− 16j) m/s2

(c) (8i) m/s2

(d) (8j) m/s2

ANS: (b)

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cee 271: applied mechanics ii, dynamics lecture 2: ch.12, sec.3h 5 · 2013. 9. 11. · 12.3...

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